Mad Scientist (Chemistry)

Heat Capacity

2007-07-05T15:59:00.001+10:00

Heat Capacity
When a substance is heated, its change in temperature(∆T) is proportional to the quantity of heat energy it absorbs(q), thus:

q/∆T = constant

The value of this constant is different for each substance and is called that substance's 'heat capacity'.

Specific Heat Capacity
The specific heat capacity(c) of a substance is the amount of heat energy required to raise 1 gram of that substance by 1 degree kelvin.

c = q/(mass * ∆T)

Molar Heat Capacity
The molar heat capacity(C) of a substance is the amount of heat energy required to raise the temperature of 1 mole of that substance by 1 degree kelvin.

C = q/(moles * ∆T)

Units
Heat energy is in joules, temperature is in degrees kelvin, and mass is in grams, thus c has units of J/(K*g) and C has units of J/(K*mol).

Enthalpy

2007-07-02T13:11:00.000+10:00

(See Energy first.)

What is enthalpy?
Enthalpy(H), or 'heat content,' is a variable that describes the thermodynamic potential(hence 'heat content') of a system.
The enthalpy of a system is defined as its internal energy plus the product of the pressure on it(P) and its volume(V).

H = E + PV

How is it used?
In many reactions the only type of work done is that of a newly formed gas expanding and pushing back the atmosphere. This is called PV work, and is equal to the product of the pressure of the atmosphere and the change in volume of the gas.

w = -P∆V

The value is negative because the gas(system) is doing work on, and thus losing energy to, the atmosphere(surroundings).

In reactions such as these, if the pressure is constant (as is most common) it is often more meaningful and convenient to examine the change in enthalpy of a system rather than the change in its internal energy. This is because by using enthalpy we can avoid having to consider PV (and thus all) work done by the system.
This is how:

H = E + PV

so at constant pressure

∆H = ∆E + P∆V

and since

∆E = q + w
&
w = -P∆V

we can see that

∆E = q - P∆V
& rearranging
q = ∆E + P∆V

so

∆H = q

As this is only true at constant pressure we write the above as

∆H = q_p

with _p effectively meaning 'at constant pressure'.

Thus to find the change in enthalpy of a system under these conditions, we only have to measure the heat transferred between the system and its surroundings.

Further
For reactions that involve little or no work, and thus where ∆E = q, knowing ∆H means knowing ∆E.

Energy

2007-06-30T23:36:00.000+10:00

When we wish to observe a change in something we can describe that thing as a 'system'. Everything else relevant to our observations of that system is it's 'surroundings'.

-The internal energy(E) of a system is the sum of the potential and kinetic energies of all its particles.

When energy is transferred between a system and its surroundings it is always transferred in the form of work and/or heat.

-Heat(q) is the energy transferred between a system and its surroundings due to a difference in their temperatures.

-Work(w) is the energy transferred when a force moves an object.

A change in internal energy of a system will be equal to the energy transferred to/from its surroundings in the form of heat and/or work. This statement is described in the equation:

∆E = q+w

where ∆ = 'change in'.

q, w, and ∆E can have positive or negative values depending on whether the system is gaining or losing energy. (loss is -, gain is +)

The 'first law of thermodynamics' or the 'law of conservation of energy' states that the total amount of energy in the universe is constant, i.e. energy cannot be created or destroyed. Any change in the energy of a system is accompanied by an opposite change of equal magnitude in its surroundings, thus there is no net gain or loss of energy.

∆Euniverse = ∆Esystem + ∆Esurroundings = 0

Nuclear Charge

2007-06-25T11:31:00.000+10:00

The nuclear charge, Z, of an atom's nucleus is the total charge of all the protons it contains (the same as its atomic number.) Lower orbital electrons can in effect "shield" higher orbital electrons from some of the positive charge of the nucleus, changing the Z that higher orbital electrons 'perceive'. Effective nuclear charge, Z_eff, is the net positive charge of the nucleus as perceived by an electron.

Periodic Table

2007-06-25T11:11:00.001+10:00

Click image for full size version.

See also:
Periodic Table Trends

The Food Chain

2007-06-25T11:01:00.000+10:00

From the Simpsons...
(I'm just testing image posting.)

Periodic Table Trends

2006-09-18T13:37:00.000+10:00

General
Atomic Size
Atomic size increases down the groups as more protons are added to the nucleus and higher electron energy levels are filled.
It decreases from left to right across the periods as the same electron energy level is filled and the larger nucleus pulls the electrons closer. (Electron shielding lowers Zeff less than increasing numbers of protons in the nucleus raises it)

Ionization Energy
Ionization Energy (IE) is the energy needed to remove an electron from an atom. It decreases with increasing atomic size because the further away from the nucleus the less Zeff.

Electronegativity
Electronegativity (EN) is the ability to attract electrons and naturally follows the same pattern as IE, however this pattern is broken with the noble gases which already have a stable electron configuration.

Groups
Group 1 (Alkali Metals)
-In their periods they have the lowest molar mass, the largest atomic size, and (thus) the lowest density.
-They only have 1 valence electron which is far from their relatively small nucleus, so their metallic bonds are relatively weak; as a result they're soft, and have low melting points.
-Low IEs make them quite reactive.

Group 2 (Alkaline Earth Metals)
-Two valence electrons and more positive charges in the nucleus lead to stronger metallic bonding; this means they're harder and have higher melting points than the group 1 metals.
-Despite their higher IE they still form +2 ions because their lattice energy is so high. (Except Be, its IE and small size only allowing it to forms covalent bonds.)
-Salts formed by ions of these elements are less soluble because of their high lattice energy.

General Properties From Group 3 Onwards
-Elements in periods containing transition and/or inner transition elements have higher IEs and smaller atomic sizes because of the greater number of protons in their nucleus.
-It is possible for these elements to lose only their p orbital valence electrons when reacting. This happens when the additional energy needed to remove the 2 s orbital valence electrons would make the energy needed for the reaction greater than that released. As a result, many of these elements can have multiple oxidation states.

Group 3
-Generally form covalent bonds but can also form ionic ones. While removing 3 electrons takes a considerable amount of energy, much energy is released when these highly charged ions form compounds, thus allowing the reaction to proceed.

Group 4
-Only Sn and Pb can form ionic compounds, however they rarely do..

Group 5
-Like period 3 they generally form covalent bonds but sometimes form ionic bonds, in this case by gaining electrons. Bi can exhist as a cation by losing only its p orbital valence electrons.

Group 6
-Form anions more readily than group 5 elements.

Group 7
-Very reactive, form anions very readily because they only need to gain 1 outer electron.
-Pure forms exhist as diatomic molecules whos melting and boiling points increase with increasing molecular weight and stronger dispersion forces going down the group.
-Reactivity decreases down the group as EN decreases.

Group 8 (Noble Gases)
-So called because they generally don't form bonds with other atoms/molecules, however they can react with elements that have very high ENs.

Diagonal Relationships
Moving down a group increases atomic size but moving right across a period decreases it. This means some elements diagonal to one another on the periodic table exhibit similar properties (eg. Li and Mg).

See also:
Periodic Table

Bridge Bonds

2006-09-18T00:13:00.000+10:00

A bridge bond is a covalent bond through which 3 atoms share 2 electrons.
For example in a B-H-B bridge bond an electron from one of the B atoms and one from the H atom would be shared between all 3 atoms. Each of the 3 atoms would consider itself the owner of the 2 electrons in terms of filling its valence shell.

Colligative Properties of Solutions

2006-09-15T00:24:00.000+10:00

(Colligative means collective.)

Properties of a solution are influenced by the number of solute particles present.
1. VP (Vapor Pressure) is lowered.
2. BP (Boiling Point) is elevated.
3. FP (Freezing Point) is lowered.
4. OP (Osmotic Pressure) is increased.

Nonvolatile Nonelectrolyte Solutions
(Volatility is the tendency of a substance to become a gas.)
(Electrolytes are a mixture of ions that when dissolved in solution conduct a current.)

When a nonvolatile nonelectrolyte substance is dissolved in solution:

1. VP is lowered.
The nonvolatile solute particles don't leave the surface of the solution, so the percentage of particles on the surface of the solution that can leave it is smaller than in the pure solvent. Particles can still enter the solution from the atmosphere at the same rate however, so the ratio of particles leaving to entering is lower than in the pure solvent.

The relationship between amount of solute and VP (known as Raoult's law) is:

VPsolvent = Xsolvent * VPpure solvent

VPsolvent = the vapor pressure of the solvent in the solution.
VPpure solvent = the vapor pressure of the pure solvent.
X= mole fraction.

If we want to know the magnitude of the change in VP we can obviously see that:

VPpure solvent - VPsolvent = ∆VP

(∆= change in)

Further:

Xsolvent + Xsolute = 1
rearranging
Xsolvent = 1 - Xsolute

thus

VPsolvent = Xsolvent * VPpure solvent = (1 - Xsolute ) * VPpure solvent

so

VPsolvent = VPpure solvent - (Xsolute * VPpure solvent)
rearranging
VPpure solvent - VPsolvent = Xsolute * VPpure solvent

so

Xsolute * VPpure solvent also = ∆VP

2. BP is elevated.
The BP of a liquid is when its VP = external pressure.
As the VP of a nonvolatile solution is lower than that of the pure solvent it needs to be heated more before the VP becomes equal to the external pressure. Thus BP is elevated.

Change in BP is directly proportional to number of solute particles.

Change in BP = Kb * m

Kb is the molal boiling point elevation constant and is specific to a given solvent, it is given in units of (degrees C/m)
m = moles of solute.

3. FP is lowered.
Similar to the reason why VP is raised. The solute particles don't solidify, therefore the ratio of particles leaving the solid unit to those entering is greater than in the pure solvent. Thus the temperature at which more particles are entering than leaving will be lower and so the FP is lower.

Change in FP = Kf * m

Kf is the molal freezing point depression constant and is specific to a given solvent, it is also given in units of (degrees C/m)

4. OP is increased.
OP is the pressure required to prevent the net movement of water from pure solvent to solution through a semipermiable membrane.

Because the solute particles cannot travel through the membrane, the amount of particles going through from the pure solvent on one side will be greater than the amount going through the other way from the solution on the other.
If the solutions on both sides are the same the OP is 0.

OP is directly proportional to the number of solute particles(n) in a given volume(V) of solution, that is, the molarity (M) of the solution.
The proportionality constant is R times the temperature (T)
So the relationship is:

OP = (nsolute/Vsolution) * RT = MRT

We can use change in VP, BP, FP and/or OP with all these relationships to find the molar mass of the solute.

Volatile Nonelectrolyte Solutions
From Raoult's law we know that:

VPsolvent = Xsolvent * VPpure solvent

If the solute is volatile the same applies to it.

VPsolute = Xsolute * VPpure solute

From the VP exerted by each gas we can find what fraction of the vapor each gas makes up.
The gas of the more volatile liquid will be a higher fraction of the vapor.

This principle is used in fractional distillation, more of the gas of the more volatile liquid is present in the vapor than the gas of the less volatile liquid. When that vapor condenses, relatively more of the more volatile liquid will be present in the solution. The more this process is repeated the more concentrated the more volatile liquid will become.

Electrolyte Solutions
Positive and negative ions in solution tend to cluster together, and this causes some of the ions to behave as if they were "tied up" (as if they were still bonded.) This reduces the effective concentration of the ions and causes the electrolyte solution to exhibit nonideal behavior.

1 mole of NaCl should dissociate into 1 mole of Na and 1 mole of Cl, but because some ions are "tied up" not all of them will affect the properties of the solution. To find how many do, we use the van't Hoff factor (i)

i = (measured value of an electrolyte solution)/(expected value of a nonelectrolyte solution)

for example:

i = (change in boiling point of a solution when 1m of NaCl is added)/(expected change in boiling point of solution if 1m of glucose was added)

in this case it's is (0.049 degrees C)/(0.026 degrees C) = 1.9

What nonelectrolyte is used doesn't matter as it is the number of dissolved particles that effect these properties and not the type.

For an electrolyte solution we would multiply the side of the equation containing the factor relating to the number of particles (m, X and M) by the number of moles of particles the electrolyte dissociates into. (NaCl into 2 for example) But since some atoms are "tied up" we multiply by i instead which gives the effective number particles affecting the solution rather than the actual number.

For example:
for NaCl instead of using

FP = 2 * (Kf * m)

we would use

FP = i * (Kf * m)
(i=1.9 in this case)

The more concentrated the solution the more ions are "tied up" and vice versa, so the more dilute the solution the closer to ideal its behavior will be.

Intermolecular Forces

2006-09-13T06:58:00.000+10:00

Ion-Dipole Forces
The attraction between an ion and a pole of a polar molecule(dipole).
BE(Bond Energy) = 40-600 kJ/mol

Dipole-Dipole Forces
The attraction between oppositely charged poles of polar molecules.
BE = 5-25 kJ/mol

Hydrogen Bond
When an H atom is bound to a small and highly electronegative atom with lone pairs (N,O or F) the electrons are drawn away from the H atom making this particular covalent bond very polar. The H atom has a relatively strong + charge and the N, O or F atom has a relatively strong - charge. The + H atom of one of these molecules forms very strong bonds with the - N, O or F atom of another of the molecules. The H atom is specifically attracted to the lone pairs of electrons of the N, O and F atoms.
BE = 10-40 kJ/mol

Charge-Induced Dipoles
Lone pair electrons are in constant motion around their atom in what we picture as a cloud of negative charge. A nearby electric field can distort the cloud by pulling it towards a positive charge or pushing it away from a negative charge. This distortion results in a temporary induced dipole in the electrons molecule.
Ion-induced dipole and dipole-induced dipole are two examples of this.
Ion-induced dipole BE = 3-15 kJ/mol
Dipole-induced dipole BE = 2-10 kJ/mol

Dispersion Forces
Momentary oscillations of electron charge can create an instantaneous dipole within an atom which in turn induces dipoles in neighboring atoms. This happens all throughout a substance so all the molecules in a substance are slightly attracted to each other.
Also known as instantaneous dipole-induced dipole forces.
BE = 0.05-40 kJ/mol

Hess's Law

2006-09-13T05:50:00.000+10:00

Hess's law of heat summation states that the enthalpy change of an overall process is the sum of the enthalpy changes of the individual steps of the process.

Because of this, we can select any number of steps with known enthalpy changes that lead us from an initial state to a final state even if in reality the process does not follow the path we have chosen.

Example
If we know how much energy it takes to heat a can of soup by 100 degrees and how much is released when it cools by 10 degrees we can work out how much energy it takes to heat it by 90 degrees. We do this by adding the enthalpy changes of heating by 100 degrees and cooling by 10.

∆H_{(heating by 90)} = ∆H_{(heating by 100)}+ ∆H_{(cooling by 10)}In reality if we heated the soup by 90 degrees we wouldn't first heat it by 100 then cool it by 10, but we can still imagine doing it this way in order to work out how much energy was needed.

(Hess' or Hess's? Hess's.)

Solutions and Solubility

2006-09-13T05:01:00.000+10:00

Like dissolves Like
When one substance dissolves in another, the intermolecular bonds in the solute must break, as must those in the solvent. The breaking of these bonds requires energy (endothermic) while the formation of new bonds between solvent and solute molecules releases energy. (exothermic)

If the strength of the bonds broken is too much greater than of those formed the substance will not dissolve.

Examples

1.
When NaCl dissolves in water the H-bonds between the water molecules break as do the ionic bonds between Na+ and Cl- ions. However the ion-dipole bonds that form between the H2O and the ions are very strong and so release much energy upn formation. This release of energy is close enough to the energy needed to break the initial bonds in the substances that the reaction can occur.

If NaCl was mixed with oil, the bonds between oil and ions would be ion-induced dipole, far too weak relative to the ionic bonding of NaCl to allow it to dissolve.

2.
When octanol is mixed with water it does not dissolve.
The dispersion forces between octanol molecules and the strong H-bonds of water must be overcome, however the bonds formed between water and octanol would only be dipole-induced dipole, too weak to offset the energy requirement for breaking the H-bonds.
If dissolved in oil however, only the disperion forces between oil molecules and those between octanol molecules would have to be overcome. The dispersion forces forming the bonds between solvent and solute would be enough that the energy released in these bonds formation would allow the reaction to occur.

(oil = any long chain length, non-polar molecule)

Extras
We can add the heats of the individual reactions together to get the overall heat of reaction thanks to Hess's Law.
If the overall reaction is exothermic, then when the solute dissolves heat will be released. This is
used in hot packs and self heating soups.
If it is endothermic it can still occur, as long as the enthalpy change is balanced by the entropy change. Such endothermic reactions are used in cold packs.

Conduction and Band Theory

2006-09-12T21:17:00.000+10:00

Band Theory
Band theory is an extension of MO(molecular orbital) theory.

The more atoms that join together in a molecule the more MOs are created.
With the number of atoms in a sample of metal (moles * Avogadro's number) a large number of MOs are formed and thus the energies of the MOs are so closely spaced that they form a continuum or band.

According to band theory the lower energy MOs are occupied by the valence electrons. These make up the valence band.
The empty MOs of higher energy make up the conduction band.

Conduction
Conductors
In conductors such as metals there is no gap between the valence band and the conduction band. This means that electrons can enter the valence band when they receive even a tiny quantity of energy. Once into an orbital in the conduction band electrons can move through the sample freely, which is why electrons in metals are considered to be completely delocalised.
When heated the movement of atoms (vibration) interferes with the flow of electrons and thus decreases the metals conductivity.

Semiconductors
In semiconductors there is an energy gap between the valence and conduction bands. This gap is relatively small however, so thermally excited electrons are able to cross it, allowing a small current to flow. For this reason heating semiconductors has the opposite effect to heating conductors in that it increases, rather than decreases, electron flow.

Insulators
In insulators the gap between bands is too large for electrons to cross, so no current is able to flow.

Doped Semiconductors
Doping semiconductors involves adding small amounts of other elements to them to increase or decrease the number of valence electrons in the bands. P has more valence electrons than Si, thus when Si is doped with P the valence electrons of the P atoms enter orbitals in the conductor band. This bridges the gap between bands and allows for some electron flow, thus increasing conductivity. This is known as an n-type semiconductor because there are extra negative charges (electrons) present.

Ga has less valence electrons than Si, thus when Si is doped with Ga some orbitals in the valence band are empty, creating positive holes. Electrons from Si atoms can migrate to these positive holes, leaving positive holes from whence they came. Conductivity has again been increased because electrons are now able to flow. This is known as a p-type semiconductor (p for positive holes).

p-n junction
A p-n junction is formed when a p-type semiconductor is placed next to an n-type. These allow current to flow through in only one direction. If the positive terminal of a battery is placed next to the p-type portion and the negative next to the n-type, the electrons from the n-type will flow across to the positive terminal and the positive holes will move towards the negative.
If the positive terminal is placed next to the n-type portion the electrons will enter directly and no current will flow across the junction. The same applies to the positive holes moving to the negative terminal.

Gases

2006-09-12T12:52:00.001+10:00

Gas Laws
Boyle's Law
At a constant temperature and pressure, the volume occupied by a fixed amount of gas is inversely proportional to the applied (external) pressure.

V = (1/P) * constant

Charles's Law
At a constant pressure the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature.

V = constant * T

Combined Gas Law
Combining the relationships in Charles's and Boyle's gives the combined gas law which is:
V = constant * (T/P)

The relationship between pressure and temperature for a gas is:

P = T * constant

(At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to the absolute temperature.)

Avogadro's Law
At fixed temperature and pressure, equal volumes of any ideal gas contain equal numbers of particles (or moles).
Thus we can see that:
At a fixed temperature and pressure, the volume occupied by a gas is directly proportional to the amount (mol) of a gas. So

V = constant * n
(n=number of moles)

The Ideal Gas Law
Combining all these relationships together we get:

PV = nRT

Where R is the universal gas constant and = 0.0821, in units of (atm*L)/(mol*K).

Extensions of the Ideal Gas Law
Finding the Density of a Gas
Because number of moles = (mass/molar mass)

n=m/M

we can rearrange the ideal gas law from

PV = nRT

to

PV = (m/M)RT

and because density(d) = m/V we can rearrange the formula to see that:

m/V = d = MP/RT

Finding Molar Mass of a Gas
Rearranging the ideal gas law again:

n = PV/RT

As n = m/M,

M= mRT/PV

or (as m/V = d)

Deviations from ideal behavior
These relationships ignore the intermolecular attractions between molecules of a gas because under normal conditions they are so small as to be insignificant. However, under very high pressures the molecules of a gas are forced closer together, so the attractions between them are much stronger and the gas' behavior is affected. The same applies at very low temperatures, where the kinetic energy of the molecules is much lower and no longer enough to overcome intermolecular attractions.